Question 32390

Aug 24, 2015

For part (b) ${K}_{s p} = 5.3 \cdot {10}^{- 6}$

Explanation:

Once again, I'll show you how to solve one example so that you can practice on the other two.

A sparingly soluble salt will not dissociate completely in aqueous solution to produce cations, which are positively charged ions, and anions, which are negatively charged ions.

Instead of a complete dissociataion, an equilibrium will be established between the solid form and the dissolved ions. In the case of calcium hydroxide, "Ca"("OH")_2, the dissociation equilibrium will look like this

${\text{Ca"("OH")_text(2(s]) rightleftharpoons "Ca"_text((aq])^(2+) + color(red)(2)"OH}}_{\textrm{\left(a q\right]}}^{-}$

Since calcium hydroxide is formed when two hydroxide anions bond ionically with one calcium cation, dissolving one mole of the solid in aqueous solution will of course produce one mole of calcium cations and two moles of hydroxide anions.

You can use an ICE table to express the equilibrium concentrations of the dissolved ions 0 here $s$ represent molar solubility

${\text{Ca"("OH")_text(2(s]) " "rightleftharpoons" " "Ca"_text((aq])^(2+)" " + " "color(red)(2)"OH}}_{\textrm{\left(a q\right]}}^{-}$

color(purple)("I")" " " " " " - " " " " " " " " " " " "0" " " " " " " " " "0
color(purple)("C")" " " " " " - " " " " " " " " " " (+s)" " " " " " "(+color(red)(2)s)
color(purple)("E")" " " " " " - " " " " " " " " " " " "s" " " " " " " " " "color(red)(2)s#

By definition, the solubility product constant for this equilibrium will be

${K}_{s p} = {\left[{\text{Ca"^(2+)] * ["OH}}^{-}\right]}^{\textcolor{red}{2}}$

which is equivalent to

${K}_{s p} = s \cdot {\left(\textcolor{red}{2} s\right)}^{\textcolor{red}{2}} = s \cdot 4 {s}^{2} = 4 {s}^{3}$

Now all you need in order to determine the value of ${K}_{s p}$ is the value of $s$, which is provided by the problem.

If $s = \text{0.011 mol/L}$, then ${K}_{s p}$ is equal to

${K}_{s p} = 4 \cdot {\left(0.011\right)}^{3} = \textcolor{g r e e n}{5.3 \cdot {10}^{- 6}}$

You can use the exact same approach to figure out the ${K}_{s p}$ for lead (II) chromate and silver phosphate, but keep in mind what the charges of the cations and anions are, they will dictate the expression of ${K}_{s p}$ in terms of $s$.